Problem 13 - Calculus Early Transcendentals

Question

Evaluate the following iterated integrals. $$\int_{0}^{\ln 2} \int_{0}^{1} 6 x e^{3 y} d x d y$$

Step-by-Step Solution

Answer

Question: Evaluate the iterated integral $$\int_{0}^{\ln 2} \int_{0}^{1} 6 x e^{3 y} d x d y$$. Answer: The value of the iterated integral is 7.

Step 1: Integrate with respect to x

First, we need to integrate the function \(6xe^{3y}\) with respect to \(x\) on the interval \([0, 1]\). Since \(y\) can be treated as a constant during this integration, we can simply integrate \(6x\): $$\int_{0}^{1} 6xe^{3y} d x = e^{3y}\int_{0}^{1} 6x dx$$ Integrating \(6x\) with respect to \(x\) gives us \(3x^2\). Now evaluate this antiderivative from 0 to 1: $$3x^2\Big|_{0}^{1} = 3(1)^2 - 3(0)^2 = 3$$ So, the inside integral becomes: $$e^{3y}(3) = 3e^{3y}$$

Step 2: Integrate with respect to y

Now, we need to integrate the resulting expression \(3e^{3y}\) with respect to \(y\) on the interval \([0, \ln 2]\): $$\int_{0}^{\ln 2} 3e^{3y} dy$$ Integrating \(3e^{3y}\) with respect to \(y\) gives us \(e^{3y}\). Evaluate this antiderivative from 0 to \(\ln 2\): $$e^{3y}\Big|_{0}^{\ln 2} = e^{3\ln 2} - e^{3 \cdot 0} = e^{\ln 8} - e^{0} = 8 - 1 = 7$$

Step 3: Final Answer

Thus, the value of the iterated integral is 7: $$\int_{0}^{\ln 2} \int_{0}^{1} 6 x e^{3 y} d x d y = 7$$